# Question #98dd6

##### 1 Answer

#### Explanation:

You know that **one mole** of any substance contains exactly *Avogadro's number*.

In your case, you're dealing with a *millimole*, which represents the

So, if you need **one mole**, it follows that you'd need **times more molecules** of carbon dioxide,

#1color(red)(cancel(color(black)("mmole"))) * (1color(red)(cancel(color(black)("mole"))))/(10^3color(red)(cancel(color(black)("mmoles")))) * (6.022 * 10^(23)"molecules CO"_2)/(1color(red)(cancel(color(black)("mole")))) = 6.022 * 10^(23)"molecules CO"_2#

You need to round this off to one sig fig, the number of sig figs you have for the number of millimoles of

#"no. of molecules of CO"_2 = color(green)(6 * 10^(23))#